cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000120 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n).

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3
Offset: 0

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Author

Keywords

Comments

The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021]
a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002
To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), ..., a(2^k-1) are the first 2^k terms, then the next 2^k terms are a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003
An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003
Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006
a(n) is the number of times n appears among the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. - Jeremy Gardiner, Jan 25 2006
a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006
The first appearance of k, k >= 0, is at a(2^k-1). - Robert G. Wilson v, Jul 27 2006
Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009
For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009
Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009
a(k*m) <= a(k) * a(m). - Robert Israel, Sep 03 2023
The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017
Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012
From Daniel Forgues, Mar 13 2015: (Start)
Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.)
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015
The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015
a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017
a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020

Examples

			Using the formula a(n) = a(floor(n / floor_pow4(n))) + a(n mod floor_pow4(n)):
  a(4) = a(1) + a(0) = 1,
  a(8) = a(2) + a(0) = 1,
  a(13) = a(3) + a(1) = 2 + 1 = 3,
  a(23) = a(1) + a(7) = 1 + a(1) + a(3) = 1 + 1 + 2 = 4.
_Gary W. Adamson_ points out (Jun 03 2009) that this can be written as a triangle:
  0,
  1,
  1,2,
  1,2,2,3,
  1,2,2,3,2,3,3,4,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
  1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
  1,2,2,3,2,3,...
where the rows converge to A063787.
From _Joerg Arndt_, Nov 09 2012: (Start)
Connection to the compositions of n as lists of parts (see comment):
[ #]:   a(n)  composition
[ 0]:   [0]   1 1 1 1 1
[ 1]:   [1]   1 1 1 2
[ 2]:   [1]   1 1 2 1
[ 3]:   [2]   1 1 3
[ 4]:   [1]   1 2 1 1
[ 5]:   [2]   1 2 2
[ 6]:   [2]   1 3 1
[ 7]:   [3]   1 4
[ 8]:   [1]   2 1 1 1
[ 9]:   [2]   2 1 2
[10]:   [2]   2 2 1
[11]:   [3]   2 3
[12]:   [2]   3 1 1
[13]:   [3]   3 2
[14]:   [3]   4 1
[15]:   [4]   5
(End)
		

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 119.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - N. J. A. Sloane, Aug 03 2012
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws. W.H. Freeman, 1991, p. 383.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The basic sequences concerning the binary expansion of n are this one, A000788, A000069, A001969, A023416, A059015, A007088.
Partial sums see A000788. For run lengths see A131534. See also A001792, A010062.
Number of 0's in n: A023416 and A080791.
a(n) = n - A011371(n).
Sum of digits of n written in bases 2-16: this sequence, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
This is Guy Steele's sequence GS(3, 4) (see A135416).
Cf. A230952 (boustrophedon transform).
Cf. A070939 (length of binary representation of n).

Programs

  • Fortran
    c See link in A139351
    
  • Haskell
    import Data.Bits (Bits, popCount)
    a000120 :: (Integral t, Bits t) => t -> Int
    a000120 = popCount
    a000120_list = 0 : c [1] where c (x:xs) = x : c (xs ++ [x,x+1])
    -- Reinhard Zumkeller, Aug 26 2013, Feb 19 2012, Jun 16 2011, Mar 07 2011
    
  • Haskell
    a000120 = concat r
        where r = [0] : (map.map) (+1) (scanl1 (++) r)
    -- Luke Palmer, Feb 16 2014
    
  • Magma
    [Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020
    
  • Magma
    [&+Intseq(n, 2):n in [0..104]]; // Marius A. Burtea, Jan 22 2020
  • Maple
    A000120 := proc(n) local w,m,i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
    A000120 := n -> add(i, i=convert(n,base,2)): # Peter Luschny, Feb 03 2011
    with(Bits): p:=n->ilog2(n-And(n,n-1)): seq(p(binomial(2*n,n)),n=0..200) # Gary Detlefs, Jan 27 2019
  • Mathematica
    Table[DigitCount[n, 2, 1], {n, 0, 105}]
    Nest[Flatten[# /. # -> {#, # + 1}] &, {0}, 7] (* Robert G. Wilson v, Sep 27 2011 *)
    Table[Plus @@ IntegerDigits[n, 2], {n, 0, 104}]
    Nest[Join[#, # + 1] &, {0}, 7] (* IWABUCHI Yu(u)ki, Jul 19 2012 *)
    Log[2, Nest[Join[#, 2#] &, {1}, 14]] (* gives 2^14 term, Carlos Alves, Mar 30 2014 *)
  • PARI
    {a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};
    
  • PARI
    {a(n) = if( n<0, 0, subst(Pol(binary(n)), x ,1))};
    
  • PARI
    {a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    a(n)=my(v=binary(n));sum(i=1,#v,v[i]) \\ Charles R Greathouse IV, Jun 24 2011
    
  • PARI
    a(n)=norml2(binary(n)) \\ better use {A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020
    
  • PARI
    a(n)=hammingweight(n) \\ Michel Marcus, Oct 19 2013
    (Common Lisp) (defun floor-to-power (n pow) (declare (fixnum pow)) (expt pow (floor (log n pow)))) (defun enabled-bits (n) (if (< n 4) (n-th n (list 0 1 1 2)) (+ (enabled-bits (floor (/ n (floor-to-power n 4)))) (enabled-bits (mod n (floor-to-power n 4)))))) ; Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
    
  • Python
    def A000120(n): return bin(n).count('1') # Chai Wah Wu, Sep 03 2014
    
  • Python
    import numpy as np
    A000120 = np.array([0], dtype="uint8")
    for bitrange in range(25): A000120 = np.append(A000120, np.add(A000120, 1))
    print([A000120[n] for n in range(0, 105)]) # Karl-Heinz Hofmann, Nov 07 2022
    
  • Python
    def A000120(n): return n.bit_count() # Requires Python 3.10 or higher. - Pontus von Brömssen, Nov 08 2022
    
  • Python
    # Also see links.
    
  • SageMath
    def A000120(n):
        if n <= 1: return Integer(n)
        return A000120(n//2) + n%2
    [A000120(n) for n in range(105)]  # Peter Luschny, Nov 19 2012
    
  • SageMath
    def A000120(n) : return sum(n.digits(2)) # Eric M. Schmidt, Apr 26 2013
    
  • Scala
    (0 to 127).map(Integer.bitCount()) // _Alonso del Arte, Mar 05 2019
    

Formula

a(0) = 0, a(2*n) = a(n), a(2*n+1) = a(n) + 1.
a(0) = 0, a(2^i) = 1; otherwise if n = 2^i + j with 0 < j < 2^i, a(n) = a(j) + 1.
G.f.: Product_{k >= 0} (1 + y*x^(2^k)) = Sum_{n >= 0} y^a(n)*x^n. - N. J. A. Sloane, Jun 04 2009
a(n) = a(n-1) + 1 - A007814(n) = log_2(A001316(n)) = 2n - A005187(n) = A070939(n) - A023416(n). - Henry Bottomley, Apr 04 2001; corrected by Ralf Stephan, Apr 15 2002
a(n) = log_2(A000984(n)/A001790(n)). - Benoit Cloitre, Oct 02 2002
For n > 0, a(n) = n - Sum_{k=1..n} A007814(k). - Benoit Cloitre, Oct 19 2002
a(n) = n - Sum_{k>=1} floor(n/2^k) = n - A011371(n). - Benoit Cloitre, Dec 19 2002
G.f.: (1/(1-x)) * Sum_{k>=0} x^(2^k)/(1+x^(2^k)). - Ralf Stephan, Apr 19 2003
a(0) = 0, a(n) = a(n - 2^floor(log_2(n))) + 1. Examples: a(6) = a(6 - 2^2) + 1 = a(2) + 1 = a(2 - 2^1) + 1 + 1 = a(0) + 2 = 2; a(101) = a(101 - 2^6) + 1 = a(37) + 1 = a(37 - 2^5) + 2 = a(5 - 2^2) + 3 = a(1 - 2^0) + 4 = a(0) + 4 = 4; a(6275) = a(6275 - 2^12) + 1 = a(2179 - 2^11) + 2 = a(131 - 2^7) + 3 = a(3 - 2^1) + 4 = a(1 - 2^0) + 5 = 5; a(4129) = a(4129 - 2^12) + 1 = a(33 - 2^5) + 2 = a(1 - 2^0) + 3 = 3. - Hieronymus Fischer, Jan 22 2006
A fixed point of the mapping 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, ... With f(i) = floor(n/2^i), a(n) is the number of odd numbers in the sequence f(0), f(1), f(2), f(3), f(4), f(5), ... - Philippe Deléham, Jan 04 2004
When read mod 2 gives the Morse-Thue sequence A010060.
Let floor_pow4(n) denote n rounded down to the next power of four, floor_pow4(n) = 4 ^ floor(log4 n). Then a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(n) = a(floor(n / floor_pow4(n))) + a(n % floor_pow4(n)). - Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
a(n) = n - Sum_{k=2..n} Sum_{j|n, j >= 2} (floor(log_2(j)) - floor(log_2(j-1))). - Hieronymus Fischer, Jun 18 2007
a(n) = A138530(n, 2) for n > 1. - Reinhard Zumkeller, Mar 26 2008
a(A077436(n)) = A159918(A077436(n)); a(A000290(n)) = A159918(n). - Reinhard Zumkeller, Apr 25 2009
a(n) = A063787(n) - A007814(n). - Gary W. Adamson, Jun 04 2009
a(n) = A007814(C(2n, n)) = 1 + A007814(C(2n-1, n)). - Vladimir Shevelev, Jul 20 2009
For odd m >= 1, a((4^m-1)/3) = a((2^m+1)/3) + (m-1)/2 (mod 2). - Vladimir Shevelev, Sep 03 2010
a(n) - a(n-1) = { 1 - a(n-1) if and only if A007814(n) = a(n-1), 1 if and only if A007814(n) = 0, -1 for all other A007814(n) }. - Brent Spillner (spillner(AT)acm.org), Sep 01 2010
a(A001317(n)) = 2^a(n). - Vladimir Shevelev, Oct 25 2010
a(n) = A139351(n) + A139352(n) = Sum_k {A030308(n, k)}. - Philippe Deléham, Oct 14 2011
From Hieronymus Fischer, Jun 10 2012: (Start)
a(n) = Sum_{j = 1..m+1} (floor(n/2^j + 1/2) - floor(n/2^j)), where m = floor(log_2(n)).
General formulas for the number of digits >= d in the base p representation of n, where 1 <= d < p: a(n) = Sum_{j = 1..m+1} (floor(n/p^j + (p-d)/p) - floor(n/p^j)), where m=floor(log_p(n)); g.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = A213629(n, 1) for n > 0. - Reinhard Zumkeller, Jul 04 2012
a(n) = A240857(n,n). - Reinhard Zumkeller, Apr 14 2014
a(n) = log_2(C(2*n,n) - (C(2*n,n) AND C(2*n,n)-1)). - Gary Detlefs, Jul 10 2014
Sum_{n >= 1} a(n)/2n(2n+1) = (gamma + log(4/Pi))/2 = A344716, where gamma is Euler's constant A001620; see Sondow 2005, 2010 and Allouche, Shallit, Sondow 2007. - Jonathan Sondow, Mar 21 2015
For any integer base b >= 2, the sum of digits s_b(n) of expansion base b of n is the solution of this recurrence relation: s_b(n) = 0 if n = 0 and s_b(n) = s_b(floor(n/b)) + (n mod b). Thus, a(n) satisfies: a(n) = 0 if n = 0 and a(n) = a(floor(n/2)) + (n mod 2). This easily yields a(n) = Sum_{i = 0..floor(log_2(n))} (floor(n/2^i) mod 2). From that one can compute a(n) = n - Sum_{i = 1..floor(log_2(n))} floor(n/2^i). - Marek A. Suchenek, Mar 31 2016
Sum_{k>=1} a(k)/2^k = 2 * Sum_{k >= 0} 1/(2^(2^k)+1) = 2 * A051158. - Amiram Eldar, May 15 2020
Sum_{k>=1} a(k)/(k*(k+1)) = A016627 = log(4). - Bernard Schott, Sep 16 2020
a(m*(2^n-1)) >= n. Equality holds when 2^n-1 >= A000265(m), but also in some other cases, e.g., a(11*(2^2-1)) = 2 and a(19*(2^3-1)) = 3. - Pontus von Brömssen, Dec 13 2020
G.f.: A(x) satisfies A(x) = (1+x)*A(x^2) + x/(1-x^2). - Akshat Kumar, Nov 04 2023

A053735 Sum of digits of (n written in base 3).

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 2, 3, 4, 1, 2, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, 3, 4, 5, 4, 5, 6, 1, 2, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, 3, 4, 5, 4, 5, 6, 3, 4, 5, 4, 5, 6, 5, 6, 7, 2, 3, 4, 3, 4, 5, 4, 5, 6, 3, 4, 5, 4, 5, 6, 5, 6, 7, 4, 5, 6, 5, 6, 7, 6, 7, 8, 1, 2, 3, 2, 3, 4, 3, 4, 5, 2, 3, 4, 3, 4, 5, 4, 5, 6, 3, 4, 5, 4, 5, 6
Offset: 0

Views

Author

Henry Bottomley, Mar 28 2000

Keywords

Comments

Also the fixed point of the morphism 0->{0,1,2}, 1->{1,2,3}, 2->{2,3,4}, etc. - Robert G. Wilson v, Jul 27 2006

Examples

			a(20) = 2 + 0 + 2 = 4 because 20 is written as 202 base 3.
From _Omar E. Pol_, Feb 20 2010: (Start)
This can be written as a triangle with row lengths A025192 (see the example in the entry A000120):
0,
1,2,
1,2,3,2,3,4,
1,2,3,2,3,4,3,4,5,2,3,4,3,4,5,4,5,6,
1,2,3,2,3,4,3,4,5,2,3,4,3,4,5,4,5,6,3,4,5,4,5,6,5,6,7,2,3,4,3,4,5,4,5,6,3,...
where the k-th row contains a(3^k+i) for 0<=i<2*3^k and converges to A173523 as k->infinity. (End) [Changed conjectures to statements in this entry. - _Franklin T. Adams-Watters_, Jul 02 2015]
G.f. = x + 2*x^2 + x^3 + 2*x^4 + 3*x^5 + 2*x^6 + 3*x^7 + 4*x^8 + x^9 + 2*x^10 + ...
		

Crossrefs

Cf. A065363, A007089, A173523. See A134451 for iterations.
Sum of digits of n written in bases 2-16: A000120, this sequence, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
Related base-3 sequences: A006047, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1), A286585, A286632, A289813, A289814.

Programs

  • Haskell
    a053735 = sum . a030341_row
    -- Reinhard Zumkeller, Feb 21 2013, Feb 19 2012
    
  • MATLAB
    m=1; for u=0:104; sol(m)=sum(dec2base(u,3)-'0'); m=m+1;end
    sol; % Marius A. Burtea, Jan 17 2019
  • Magma
    [&+Intseq(n,3):n in [0..104]]; // Marius A. Burtea, Jan 17 2019
    
  • Maple
    seq(convert(convert(n,base,3),`+`),n=0..100); # Robert Israel, Jul 02 2015
  • Mathematica
    Table[Plus @@ IntegerDigits[n, 3], {n, 0, 100}] (* or *)
    Nest[Join[#, # + 1, # + 2] &, {0}, 6] (* Robert G. Wilson v, Jul 27 2006 and modified Jul 27 2014 *)
  • PARI
    {a(n) = if( n<1, 0, a(n\3) + n%3)}; /* Michael Somos, Mar 06 2004 */
    
  • PARI
    A053735(n)=sumdigits(n,3) \\ Requires version >= 2.7. Use sum(i=1,#n=digits(n,3),n[i]) in older versions. - M. F. Hasler, Mar 15 2016
    
  • Scheme
    (define (A053735 n) (let loop ((n n) (s 0)) (if (zero? n) s (let ((d (mod n 3))) (loop (/ (- n d) 3) (+ s d)))))) ;; For R6RS standard. Use modulo instead of mod in older Schemes like MIT/GNU Scheme. - Antti Karttunen, Jun 03 2017
    

Formula

From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(3n) = a(n), a(3n + 1) = a(n) + 1, a(3n + 2) = a(n) + 2.
a(n) = n - 2*Sum_{k>0} floor(n/3^k) = n - 2*A054861(n). (End)
a(n) = A062756(n) + 2*A081603(n). - Reinhard Zumkeller, Mar 23 2003
G.f.: (Sum_{k >= 0} (x^(3^k) + 2*x^(2*3^k))/(1 + x^(3^k) + x^(2*3^k)))/(1 - x). - Michael Somos, Mar 06 2004, corrected by Franklin T. Adams-Watters, Nov 03 2005
In general, the sum of digits of (n written in base b) has generating function (Sum_{k>=0} (Sum_{0 <= i < b} i*x^(i*b^k))/(Sum_{i=0..b-1} x^(i*b^k)))/(1-x). - Franklin T. Adams-Watters, Nov 03 2005
First differences of A094345. - Vladeta Jovovic, Nov 08 2005
a(A062318(n)) = n and a(m) < n for m < A062318(n). - Reinhard Zumkeller, Feb 26 2008
a(n) = A138530(n,3) for n > 2. - Reinhard Zumkeller, Mar 26 2008
a(n) <= 2*log_3(n+1). - Vladimir Shevelev, Jun 01 2011
a(n) = Sum_{k>=0} A030341(n, k). - Philippe Deléham, Oct 21 2011
G.f. satisfies G(x) = (x+2*x^2)/(1-x^3) + (1+x+x^2)*G(x^3), and has a natural boundary at |x|=1. - Robert Israel, Jul 02 2015
a(n) = A056239(A006047(n)). - Antti Karttunen, Jun 03 2017
a(n) = A000120(A289813(n)) + 2*A000120(A289814(n)). - Antti Karttunen, Jul 20 2017
a(0) = 0; a(n) = a(n - 3^floor(log_3(n))) + 1. - Ilya Gutkovskiy, Aug 23 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 3*log(3)/2 (Shallit, 1984). - Amiram Eldar, Jun 03 2021

A194959 Fractalization of (1 + floor(n/2)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 3, 4, 2, 1, 3, 5, 4, 2, 1, 3, 5, 6, 4, 2, 1, 3, 5, 7, 6, 4, 2, 1, 3, 5, 7, 8, 6, 4, 2, 1, 3, 5, 7, 9, 8, 6, 4, 2, 1, 3, 5, 7, 9, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 12, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 11, 13, 12, 10, 8, 6, 4, 2, 1, 3, 5
Offset: 1

Views

Author

Clark Kimberling, Sep 06 2011

Keywords

Comments

Suppose that p(1), p(2), p(3), ... is an integer sequence satisfying 1 <= p(n) <= n for n >= 1. Define g(1)=(1) and for n > 1, form g(n) from g(n-1) by inserting n so that its position in the resulting n-tuple is p(n). The sequence f obtained by concatenating g(1), g(2), g(3), ... is clearly a fractal sequence, here introduced as the fractalization of p. The interspersion associated with f is here introduced as the interspersion fractally induced by p, denoted by I(p); thus, the k-th term in the n-th row of I(p) is the position of the k-th n in f. Regarded as a sequence, I(p) is a permutation of the positive integers; its inverse permutation is denoted by Q(p).
...
Example: Let p=(1,2,2,3,3,4,4,5,5,6,6,7,7,...)=A008619. Then g(1)=(1), g(2)=(1,2), g(3)=(1,3,2), so that
f=(1,1,2,1,3,2,1,3,4,2,1,3,5,4,2,1,3,5,6,4,2,1,...)=A194959; and I(p)=A057027, Q(p)=A064578.
The interspersion I(P) has the following northwest corner, easily read from f:
1 2 4 7 11 16 22
3 6 10 15 21 28 36
5 8 12 17 23 30 38
9 14 20 27 35 44 54
...
Following is a chart of selected p, f, I(p), and Q(p):
p f I(p) Q(p)
Count odd numbers up to n, then even numbers down from n. - Franklin T. Adams-Watters, Jan 21 2012
This sequence defines the square array A(n,k), n > 0 and k > 0, read by antidiagonals and the triangle T(n,k) = A(n+1-k,k) for 1 <= k <= n read by rows (see Formula and Example). - Werner Schulte, May 27 2018

Examples

			The sequence p=A008619 begins with 1,2,2,3,3,4,4,5,5,..., so that g(1)=(1). To form g(2), write g(1) and append 2 so that in g(2) this 2 has position p(2)=2: g(2)=(1,2). Then form g(3) by inserting 3 at position p(3)=2: g(3)=(1,3,2), and so on. The fractal sequence A194959 is formed as the concatenation g(1)g(2)g(3)g(4)g(5)...=(1,1,2,1,3,2,1,3,4,2,1,3,5,4,2,...).
From _Werner Schulte_, May 27 2018: (Start)
This sequence seen as a square array read by antidiagonals:
  n\k: 1  2  3  4  5   6   7   8   9  10  11  12 ...
  ===================================================
   1   1  2  2  2  2   2   2   2   2   2   2   2 ... (see A040000)
   2   1  3  4  4  4   4   4   4   4   4   4   4 ... (see A113311)
   3   1  3  5  6  6   6   6   6   6   6   6   6 ...
   4   1  3  5  7  8   8   8   8   8   8   8   8 ...
   5   1  3  5  7  9  10  10  10  10  10  10  10 ...
   6   1  3  5  7  9  11  12  12  12  12  12  12 ...
   7   1  3  5  7  9  11  13  14  14  14  14  14 ...
   8   1  3  5  7  9  11  13  15  16  16  16  16 ...
   9   1  3  5  7  9  11  13  15  17  18  18  18 ...
  10   1  3  5  7  9  11  13  15  17  19  20  20 ...
  etc.
This sequence seen as a triangle read by rows:
  n\k:  1  2  3  4  5   6   7   8   9  10  11  12  ...
  ======================================================
   1    1
   2    1  2
   3    1  3  2
   4    1  3  4  2
   5    1  3  5  4  2
   6    1  3  5  6  4   2
   7    1  3  5  7  6   4   2
   8    1  3  5  7  8   6   4   2
   9    1  3  5  7  9   8   6   4   2
  10    1  3  5  7  9  10   8   6   4   2
  11    1  3  5  7  9  11  10   8   6   4   2
  12    1  3  5  7  9  11  12  10   8   6   4   2
  etc.
(End)
		

References

  • Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168.

Crossrefs

Cf. A000142, A000217, A005408, A005843, A008619, A057027, A064578, A209229, A210535, A219977; A000012 (col 1), A157532 (col 2), A040000 (row 1), A113311 (row 2); A194029 (introduces the natural fractal sequence and natural interspersion of a sequence - different from those introduced at A194959).
Cf. A003558 (g permutation order), A102417 (index), A330081 (on bits), A057058 (inverse).

Programs

  • Mathematica
    r = 2; p[n_] := 1 + Floor[n/r]
    Table[p[n], {n, 1, 90}]  (* A008619 *)
    g[1] = {1}; g[n_] := Insert[g[n - 1], n, p[n]]
    f[1] = g[1]; f[n_] := Join[f[n - 1], g[n]]
    f[20] (* A194959 *)
    row[n_] := Position[f[30], n];
    u = TableForm[Table[row[n], {n, 1, 5}]]
    v[n_, k_] := Part[row[n], k];
    w = Flatten[Table[v[k, n - k + 1], {n, 1, 13},
    {k, 1, n}]]  (* A057027 *)
    q[n_] := Position[w, n]; Flatten[
    Table[q[n], {n, 1, 80}]]  (* A064578 *)
    Flatten[FoldList[Insert[#1, #2, Floor[#2/2] + 1] &, {}, Range[10]]] (* Birkas Gyorgy, Jun 30 2012 *)
  • PARI
    T(n,k) = min(k<<1-1,(n-k+1)<<1); \\ Kevin Ryde, Oct 09 2020

Formula

From Werner Schulte, May 27 2018 and Jul 10 2018: (Start)
Seen as a triangle: It seems that the triangle T(n,k) for 1 <= k <= n (see Example) is the mirror image of A210535.
Seen as a square array A(n,k) and as a triangle T(n,k):
A(n,k) = 2*k-1 for 1 <= k <= n, and A(n,k) = 2*n for 1 <= n < k.
A(n+1,k+1) = A(n,k+1) + A(n,k) - A(n-1,k) for k > 0 and n > 1.
A(n,k) = A(k,n) - 1 for n > k >= 1.
P(n,x) = Sum_{k>0} A(n,k)*x^(k-1) = (1-x^n)*(1-x^2)/(1-x)^3 for n >= 1.
Q(y,k) = Sum_{n>0} A(n,k)*y^(n-1) = 1/(1-y) for k = 1 and Q(y,k) = Q(y,1) + P(k-1,y) for k > 1.
G.f.: Sum_{n>0, k>0} A(n,k)*x^(k-1)*y^(n-1) = (1+x)/((1-x)*(1-y)*(1-x*y)).
Sum_{k=1..n} A(n+1-k,k) = Sum_{k=1..n} T(n,k) = A000217(n) for n > 0.
Sum_{k=1..n} (-1)^(k-1) * A(n+1-k,k) = Sum_{k=1..n} (-1)^(k-1) * T(n,k) = A219977(n-1) for n > 0.
Product_{k=1..n} A(n+1-k,k) = Product_{k=1..n} T(n,k) = A000142(n) for n > 0.
A(n+m,n) = A005408(n-1) for n > 0 and some fixed m >= 0.
A(n,n+m) = A005843(n) for n > 0 and some fixed m > 0.
Let A_m be the upper left part of the square array A(n,k) with m rows and m columns. Then det(A_m) = 1 for some fixed m > 0.
The P(n,x) satisfy the recurrence equation P(n+1,x) = P(n,x) + x^n*P(1,x) for n > 0 and initial value P(1,x) = (1+x)/(1-x).
Let B(n,k) be multiplicative with B(n,p^e) = A(n,e+1) for e >= 0 and some fixed n > 0. That yields the Dirichlet g.f.: Sum_{k>0} B(n,k)/k^s = (zeta(s))^3/(zeta(2*s)*zeta(n*s)).
Sum_{k=1..n} A(k,n+1-k)*A209229(k) = 2*n-1. (conjectured)
(End)
From Kevin Ryde, Oct 09 2020: (Start)
T(n,k) = 2*k-1 if 2*k-1 <= n, or 2*(n+1-k) if 2*k-1 > n. [Lévy, chapter 1 section 1 equations (a),(b)]
Fixed points T(n,k)=k for k=1 and k = (2/3)*(n+1) when an integer. [Lévy, chapter 1 section 2 equation (3)]
(End)

Extensions

Name corrected by Franklin T. Adams-Watters, Jan 21 2012

A008611 a(n) = a(n-3) + 1, with a(0)=a(2)=1, a(1)=0.

Original entry on oeis.org

1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 7, 6, 7, 8, 7, 8, 9, 8, 9, 10, 9, 10, 11, 10, 11, 12, 11, 12, 13, 12, 13, 14, 13, 14, 15, 14, 15, 16, 15, 16, 17, 16, 17, 18, 17, 18, 19, 18, 19, 20, 19, 20, 21, 20, 21, 22, 21, 22, 23, 22, 23, 24, 23, 24, 25, 24, 25, 26, 25, 26, 27, 26, 27, 28
Offset: 0

Views

Author

N. J. A. Sloane, Mar 15 1996

Keywords

Comments

Molien series of 2-dimensional representation of cyclic group of order 3 over GF(2).
One step back, two steps forward.
The crossing number of the graph C(n, {1,3}), n >= 8, is [n/3] + n mod 3, which gives this sequence starting at the first 4. [Yang Yuansheng et al.]
A Chebyshev transform of A078008. The g.f. is the image of (1-x)/(1-x-2*x^2) (g.f. of A078008) under the Chebyshev transform A(x)-> (1/(1+x^2))*A(x/(1+x^2)). - Paul Barry, Oct 15 2004
A047878 is an essentially identical sequence. - Anton Chupin, Oct 24 2009
Rhyme scheme of Dante Alighieri's "Divine Comedy." - David Gaita, Feb 11 2011
A194960 results from deleting the first four terms of A008611. Note that deleting the first term or first four terms of A008611 leaves a concatenation of segments (n, n+1, n+2); for related concatenations, see
A008619, (n,n+1) after deletion of first term;
A053737, (n,n+1,n+2,n+3) beginning with n=0;
A053824, (n to n+4) beginning with n=0. - Clark Kimberling, Sep 07 2011
It appears that a(n) is the number of roots of x^(n+1) + x + 1 inside the unit circle. - Michel Lagneau, Nov 02 2012
Also apparently for n >= 2: a(n) is the largest remainder r that results from dividing n+2 by 1..n+2 more than once, i.e., a(n) = max(i, A072528(n+2,i)>1). - Ralf Stephan, Oct 21 2013
Number of n-element subsets of [n+1] whose sum is a multiple of 3. a(4) = 1: {1,2,4,5}. - Alois P. Heinz, Feb 06 2017
It appears that a(n) is the number of roots of the Fibonacci polynomial F(n+2,x) strictly inside the unit circle of the complex plane. - Michel Lagneau, Apr 07 2017
For the proof of the preceding conjecture see my comments under A008615 and A049310. Chebyshev S(n,x) = i^n*F(n+1,-i*x), with i = sqrt(-1). - Wolfdieter Lang, May 06 2017
The sequence is the interleaving of three sequences: the positive integers (A000027), the nonnegative integers (A001477), and the positive integers, in that order. - Guenther Schrack, Nov 07 2020
a(n) is the number of multiples of 3 between n and 2n. - Christian Barrientos, Dec 20 2021
a(n) is the least number of football games a team has to play to be able to get n-1 points, where a win is 3 points, a draw is 1 point, and a loss is 0 points. - Sigurd Kittilsen, Dec 01 2022

Examples

			G.f. = 1 + x^2 + 2*x^3 + x^4 + 2*x^5 + 3*x^6 + 2*x^7 + 3*x^8 + 4*x^9 + ...
		

References

  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 103.

Crossrefs

Programs

  • Haskell
    a008611 n = n' + mod r 2 where (n', r) = divMod (n + 1) 3
    a008611_list = f [1,0,1] where f xs = xs ++ f (map (+ 1) xs)
    -- Reinhard Zumkeller, Nov 25 2013
    
  • Magma
    [(n-1)-2*Floor((n-1)/3): n in [0..90]]; // Vincenzo Librandi, Aug 21 2011
    
  • Maple
    with(numtheory): for n from 1 to 70 do:it:=0:
    y:=[fsolve(x^n+x+1, x, complex)] : for m from 1 to nops(y) do : if abs(y[m])< 1 then it:=it+1:else fi:od: printf(`%d, `,it):od:
    A008611:=n->(n-1)-2*floor((n-1)/3); seq(A008611(n), n=0..50); # Wesley Ivan Hurt, May 18 2014
  • Mathematica
    With[{nn=30},Riffle[Riffle[Range[nn],Range[0,nn-1]],Range[nn],3]] (* or *) RecurrenceTable[{a[0]==a[2]==1,a[1]==0,a[n]==a[n-3]+1},a,{n,90}] (* Harvey P. Dale, Nov 06 2011 *)
    LinearRecurrence[{1, 0, 1, -1}, {1, 0, 1, 2}, 100] (* Vladimir Joseph Stephan Orlovsky, Feb 23 2012 *)
    a[ n_] := Quotient[n - 1, 3] + Mod[n + 2, 3]; (* Michael Somos, Jan 23 2014 *)
  • PARI
    {a(n) = (n-1) \ 3 + (n+2) % 3}; /* Michael Somos, Jan 23 2014 */

Formula

a(n) = a(n-3) + 1.
a(n) = (n-1) - 2*floor((n-1)/3).
G.f.: (1 + x^2 + x^4)/(1 - x^3)^2.
After the initial term, has form {n, n+1, n+2} for n=0, 1, 2, ...
From Paul Barry, Mar 18 2004: (Start)
a(n) = Sum_{k=0..n} (-1)^floor(2*(k-2)/3);
a(n) = 4*sqrt(3)*cos(2*Pi*n/3 + Pi/6)/9 + (n+1)/3. (End)
From Paul Barry, Oct 15 2004: (Start)
G.f.: (1 - x + x^2)/((1 + x + x^2)*(x-1)^2);
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*A078008(n-2k)*(-1)^k. (End)
a(n) = -a(-2-n) for all n in Z.
Euler transform of length 6 sequence [0, 1, 2, 0, 0, -1]. - Michael Somos, Jan 23 2014
a(n) = ((n-1) mod 3) + floor((n-1)/3). - Wesley Ivan Hurt, May 18 2014
PSUM transform of A257075. - Michael Somos, Apr 15 2015
a(n) = A194960(n-3), n >= 0, with extended A194960. See the a(n) formula two lines above. - Wolfdieter Lang, May 06 2017
From Guenther Schrack, Nov 07 2020: (Start)
a(n) = (3*n + 3 + 2*(w^(2*n)*(1 - w) + w^n*(2 + w)))/9, where w = (-1 + sqrt(-3))/2, a primitive third root of unity;
a(n) = (n + 1 + 2*A049347(n))/3;
a(n) = (2*n - A330396(n-1))/3. (End)
E.g.f.: (3*exp(x)*(1 + x) + exp(-x/2)*(6*cos(sqrt(3)*x/2) - 2*sqrt(3)*sin(sqrt(3)*x/2)))/9. - Stefano Spezia, May 06 2022
Sum_{n>=2} (-1)^n/a(n) = 3*log(2) - 1. - Amiram Eldar, Sep 10 2023

A053824 Sum of digits of (n written in base 5).

Original entry on oeis.org

0, 1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 7, 8, 9, 10, 11, 4, 5, 6
Offset: 0

Views

Author

Henry Bottomley, Mar 28 2000

Keywords

Comments

Also the fixed point of the morphism 0->{0,1,2,3,4}, 1->{1,2,3,4,5}, 2->{2,3,4,5,6}, etc. - Robert G. Wilson v, Jul 27 2006

Examples

			a(20) = 4 + 0 = 4 because 20 is written as 40 in base 5.
From _Omar E. Pol_, Feb 21 2010: (Start)
It appears that this can be written as a triangle:
  0,
  1,2,3,4,
  1,2,3,4,5,2,3,4,5,6,3,4,5,6,7,4,5,6,7,8,
  1,2,3,4,5,2,3,4,5,6,3,4,5,6,7,4,5,6,7,8,5,6,7,8,9,2,3,4,5,6,3,4,5,6,7,4,5,...
See the conjecture in the entry A000120. (End)
		

Crossrefs

Sum of digits of n written in bases 2-16: A000120, A053735, A053737, this sequence, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
Cf. A173525. - Omar E. Pol, Feb 21 2010
Cf. A173670 (last nonzero decimal digit of (10^n)!). - Washington Bomfim, Jan 01 2011

Programs

  • Haskell
    a053824 0 = 0
    a053824 x = a053824 x' + d  where (x', d) = divMod x 5
    -- Reinhard Zumkeller, Jan 31 2014
    
  • Magma
    [&+Intseq(n, 5):n in [0..100]]; // Marius A. Burtea, Aug 24 2019
  • Mathematica
    Table[Plus @@ IntegerDigits[n, 5], {n, 0, 100}] (* or *)
    Nest[Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 4}]] &, {0}, 4] (* Robert G. Wilson v, Jul 27 2006 *)
    f[n_] := n - 4 Sum[Floor[n/5^k], {k, n}]; Array[f, 103, 0]
  • PARI
    a(n)=if(n<1,0,if(n%5,a(n-1)+1,a(n/5)))
    
  • PARI
    a(n) = sumdigits(n, 5); \\ Michel Marcus, Aug 24 2019
    

Formula

From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(5n+i) = a(n) + i for 0 <= i <= 4;
a(n) = n - 4*Sum_{k>=1} floor(n/5^k) = n - 4*A027868(n). (End)
a(n) = A138530(n,5) for n > 4. - Reinhard Zumkeller, Mar 26 2008
If i >= 2, a(2^i) mod 4 = 0. - Washington Bomfim, Jan 01 2011
a(n) = Sum_{k>=0} A031235(n,k). - Philippe Deléham, Oct 21 2011
a(0) = 0; a(n) = a(n - 5^floor(log_5(n))) + 1. - Ilya Gutkovskiy, Aug 23 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 5*log(5)/4 (Shallit, 1984). - Amiram Eldar, Jun 03 2021

A053830 Sum of digits of (n written in base 9).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 5, 6, 7, 8, 9, 10, 11, 12, 5, 6, 7, 8, 9, 10, 11, 12, 13, 6, 7, 8, 9, 10, 11, 12, 13, 14, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1, 2, 3, 4, 5, 6, 7, 8, 9
Offset: 0

Views

Author

Henry Bottomley, Mar 28 2000

Keywords

Comments

Also the fixed point of the morphism 0->{0,1,2,3,4,5,6,7,8}, 1->{1,2,3,4,5,6,7,8,9}, 2->{2,3,4,5,6,7,8,9,10}, etc. - Robert G. Wilson v, Jul 27 2006

Examples

			a(20) = 2+2 = 4 because 20 is written as 22 base 9.
From _Omar E. Pol_, Feb 23 2010: (Start)
It appears that this can be written as a triangle (see the conjecture in the entry A000120):
0;
1,2,3,4,5,6,7,8;
1,2,3,4,5,6,7,8,9,2,3,4,5,6,7,8,9,10,3,4,5,6,7,8,9,10,11,4,5,6,7,8,9,10,11,...
where the rows converge to A173529. (End)
		

Crossrefs

Programs

  • Magma
    [&+Intseq(n, 9):n in [0..100]]; // Marius A. Burtea, Aug 24 2019
  • Mathematica
    Table[Plus @@ IntegerDigits[n, 9], {n, 0, 100}] (* or *)
    Nest[ Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 8}]] &, {0}, 3] (* Robert G. Wilson v, Jul 27 2006 *)
  • PARI
    a(n)=if(n<1,0,if(n%9,a(n-1)+1,a(n/9)))
    

Formula

From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(9n+i) = a(n) + i for 0 <= i <= 8;
a(n) = n - 8*Sum_{k>=1} floor(n/9^k) = n - 8*A054898(n). (End)
a(n) = A138530(n,9) for n > 8. - Reinhard Zumkeller, Mar 26 2008
a(n) = Sum_{k>=0} A031087(n,k). - Philippe Deléham, Oct 21 2011
a(0) = 0; a(n) = a(n - 9^floor(log_9(n))) + 1. - Ilya Gutkovskiy, Aug 24 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 9*log(9)/8 (Shallit, 1984). - Amiram Eldar, Jun 03 2021

A053827 Sum of digits of (n written in base 6).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 6, 7, 8, 9, 10, 11, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7, 8, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 6, 7, 8, 9, 10
Offset: 0

Views

Author

Henry Bottomley, Mar 28 2000

Keywords

Comments

Also the fixed point of the morphism 0->{0,1,2,3,4,5}, 1->{1,2,3,4,5,6}, 2->{2,3,4,5,6,7}, etc. - Robert G. Wilson v, Jul 27 2006
Sum of six consecutive terms is (15,21,27,33,39,45; 21,27,33,39,45,51; 27,33,39,45,51,57; and so on). - Vincenzo Librandi, Aug 02 2010

Examples

			a(20)=3+2=5 because 20 is written as 32 base 6.
From _Omar E. Pol_, Feb 21 2010: (Start)
It appears that this can be written as a triangle :
  0,
  1,2,3,4,5,
  1,2,3,4,5,6,2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,
  1,2,3,4,5,6,2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,2...
where the rows converge to A173526.
See the conjecture in the entry A000120. (End)
		

Crossrefs

Sum of digits of n written in bases 2-16: A000120, A053735, A053737, A053824, this sequence, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.
Cf. A173526. - Omar E. Pol, Feb 21 2010

Programs

  • Magma
    [&+Intseq(n,6):n in [0..105]]; // Marius A. Burtea, Aug 24 2019
  • Mathematica
    Table[Plus @@ IntegerDigits[n, 6], {n, 0, 100}] (* or *)
    Nest[ Flatten[ #1 /. a_Integer -> Table[a + i, {i, 0, 5}]] &, {0}, 4] (* Robert G. Wilson v, Jul 27 2006 *)
  • PARI
    a(n)=if(n<1,0,if(n%6,a(n-1)+1,a(n/6)))
    
  • PARI
    a(n) = sumdigits(n, 6); \\ Michel Marcus, Aug 24 2019
    

Formula

From Benoit Cloitre, Dec 19 2002: (Start)
a(0) = 0, a(6n+i) = a(n)+i for 0 <= i <= 5.
a(n) = n-5*(Sum_{k>0} floor(n/6^k)) = n-5*A054895(n). (End)
a(n) = A138530(n,6) for n > 5. - Reinhard Zumkeller, Mar 26 2008
a(n) = Sum_{k>=0} A030567(n,k). - Philippe Deléham, Oct 21 2011
a(0) = 0; a(n) = a(n - 6^floor(log_6(n))) + 1. - Ilya Gutkovskiy, Aug 23 2019
Sum_{n>=1} a(n)/(n*(n+1)) = 6*log(6)/5 (Shallit, 1984). - Amiram Eldar, Jun 03 2021

A010064 Base 4 self or Colombian numbers (not of form k + sum of base 4 digits of k).

Original entry on oeis.org

1, 3, 8, 13, 18, 20, 25, 30, 35, 37, 42, 47, 52, 54, 59, 64, 73, 78, 83, 85, 90, 95, 100, 102, 107, 112, 117, 119, 124, 129, 138, 143, 148, 150, 155, 160, 165, 167, 172, 177, 182, 184, 189, 194, 203, 208, 213, 215, 220, 225, 230, 232, 237, 242, 247, 249, 254
Offset: 1

Views

Author

Keywords

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.24, pp. 179-180.
  • József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, pp. 384-386.

Crossrefs

Programs

  • Mathematica
    s[n_] := n + Plus @@ IntegerDigits[n, 4]; m = 250; Complement[Range[m], Array[s, m]] (* Amiram Eldar, Nov 28 2020 *)

A138530 Triangle read by rows: T(n,k) = sum of digits of n in base k representation, 1<=k<=n.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 2, 1, 5, 2, 3, 2, 1, 6, 2, 2, 3, 2, 1, 7, 3, 3, 4, 3, 2, 1, 8, 1, 4, 2, 4, 3, 2, 1, 9, 2, 1, 3, 5, 4, 3, 2, 1, 10, 2, 2, 4, 2, 5, 4, 3, 2, 1, 11, 3, 3, 5, 3, 6, 5, 4, 3, 2, 1, 12, 2, 2, 3, 4, 2, 6, 5, 4, 3, 2, 1, 13, 3, 3, 4, 5, 3, 7, 6, 5, 4, 3, 2, 1, 14, 3, 4, 5, 6, 4, 2, 7, 6, 5, 4, 3, 2, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Mar 26 2008

Keywords

Comments

A131383(n) = sum of n-th row;
A000027(n) = T(n,1);
A000120(n) = T(n,2) for n>1;
A053735(n) = T(n,3) for n>2;
A053737(n) = T(n,4) for n>3;
A053824(n) = T(n,5) for n>4;
A053827(n) = T(n,6) for n>5;
A053828(n) = T(n,7) for n>6;
A053829(n) = T(n,8) for n>7;
A053830(n) = T(n,9) for n>8;
A007953(n) = T(n,10) for n>9;
A053831(n) = T(n,11) for n>10;
A053832(n) = T(n,12) for n>11;
A053833(n) = T(n,13) for n>12;
A053834(n) = T(n,14) for n>13;
A053835(n) = T(n,15) for n>14;
A053836(n) = T(n,16) for n>15;
A007395(n) = T(n,n-1) for n>1;
A000012(n) = T(n,n).

Examples

			Start of the triangle for n in base k representation:
......................1
....................11....10
......... ........111....11...10
................1111...100...11..10
..............11111...101...12..11..10
............111111...110...20..12..11..10
..........1111111...111...21..13..12..11..10
........11111111..1000...22..20..13..12..11..10
......111111111..1001..100..21..14..13..12..11..10
....1111111111..1010..101..22..20..14..13..12..11..10
..11111111111..1011..102..23..21..15..14..13..12..11..10
111111111111..1100..110..30..22..20..15..14..13..12..11..10,
and the triangle of sums of digits starts:
......................1
.....................2...1
......... ..........3...2...1
...................4...1...2...1
..................5...2...3...2...1
.................6...2...2...3...2...1
................7...3...3...4...3...2...1
...............8...1...4...2...4...3...2...1
..............9...2...1...3...5...4...3...2...1
............10...2...2...4...2...5...4...3...2...1
...........11...3...3...5...3...6...5...4...3...2...1
..........12...2...2...3...4...2...6...5...4...3...2...1.
		

Crossrefs

Cf. A007953. See A240236 for another version.
Cf. A002260.

Programs

  • Haskell
    a138530 n k = a138530_tabl !! (n-1) !! (k-1)
    a138530_row n = a138530_tabl !! (n-1)
    a138530_tabl = zipWith (map . flip q) [1..] a002260_tabl where
       q 1 n = n
       q b n = if n < b then n else q b n' + d where (n', d) = divMod n b
    -- Reinhard Zumkeller, Apr 29 2015
  • Mathematica
    T[n_, k_] := If[k == 1, n, Total[IntegerDigits[n, k]]];
    Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 25 2021 *)

A230638 Smallest number m such that u + (sum of base-4 digits of u) = m has exactly n solutions.

Original entry on oeis.org

0, 17, 16385, 16777234
Offset: 1

Views

Author

N. J. A. Sloane, Oct 31 2013

Keywords

Comments

Indices of records in A230632: a(n) is the index of the first n in A230632.
The terms are a(1)=0, a(2)=4^2+1, a(3)=4^7+1, a(4)=4^12+17+1, a(5)=4^5368+17+1, a(6)=4^10924+16385+1, a(7)=4^5597880+16385+20. Note that a(7) breaks the pattern of the first six terms.
a(8) = 4^16777229 + 4^12 + 19.
For the leading power of 4 see A230637.

Examples

			n=2: the two solutions to u+(base-4 digit-sum of u) = 17 are 13 and 16.
n=3: the three solutions to u+(base-4 digit-sum of u) = 4^7+1 are 4^7, 4^7-15, 4^7-18.
n=4: the four solutions to u+(base-4 digit-sum of u) = 4^12+17+1 are 4^12+{16, 13, -14, -17}.
		

Crossrefs

Cf. A230637.
Related base-4 sequences: A053737, A230631, A230632, A010064, A230633, A230634, A230635, A230636, A230637, A230638, A010065 (trajectory of 1)
Smallest number m such that u + (sum of base-b digits of u) = m has exactly n solutions, for bases 2 through 10: A230303, A230640, A230638, A230867, A238840, A238841, A238842, A238843, A006064.

Extensions

a(8) from Max Alekseyev, Oct 31 2013
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